Optimization
Methods
in
Management
Science
MIT
15.053
Recitation
7
TAs:
Giacomo
Nannicini,
Ebrahim
Nasrabadi
Problem
1
Suppose
branch
and
bound
is
being
applied
to
a
0–1
integer
program
in
which
we
are
maximizing.
By
node
11
the
first
three
variables
have
been
fixed.
An
incumbent
has
already
been
found
(in
another
part
of
the
tree)
and
has
value
15.
Furthermore,
x
4
has
been
chosen
to
be
the
next
variable
to
branch
on.
This
situation
is
illustrated
in
the
following
fragment
of
the
branch
and
bound
tree:
11
12
13
Incumbent: z = 15
z
LP
°
16
x
4
= 1
x
4
= 0
x
1
,
x
2
,
x
3
fixed
Figure
1:
A
fragment
of
the
branch
and
bound
tree
for
Problem
1.
The
LP
relaxation
solved
at
node
11
(which
has
x
4
and
x
5
as
free
variables)
was:
max
18
x
4
+
ax
5
+
10
s.t.
8
x
4
+
10
x
5
≤
5
,
0
≤
x
4
≤
1
,
0
≤
x
5
≤
1
,
(Hint:
For
the
following
questions,
the
LP
relaxation
essentially
becomes
a
one
variable
problem
that
can
be
solved
by
inspection.)
Part
A.1
Suppose
x
4
is
set
to
1.
Does
node
12
get
pruned
(fathomed)?
Justify
your
answer.
Solution.
Yes,
node
12
gets
fathomed.
If
x4
is
set
to
1,
then
the
LP
relaxation
becomes:
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�
max
ax
5
+
28
s.t.
10
x
5
≤ −
3
,
0
≤
x
5
≤
1
,
The
relaxation
is
infeasible
because
of
the
first
constraint,
so
node
12
gets
fathomed.
Part
B.1
Suppose
x
4
is
set
to
0
and
parameter
a
=
12.
Does
node
13
get
pruned
(fathomed)?
Justify
your
answer.
Solution.
No,
node
13
does
not
get
fathomed.
If
x
4
=
0
and
a
=
12,
then
the
LP
relaxation
becomes:
max
12
x
5
+
10
s.t.
10
x
5
≤
5
,
0
≤
x
5
≤
1
,
The
optimal
solution
to
this
relaxation
is
x
5
= 1
/
2,
with
an
objective
value
of
16.
Because
the
solution
is
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 Spring '07
 JamesOrli
 Management, Linear Programming, Optimization, LP, Approximation algorithm, Cutting stock problem

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